n identical calls are joined in series with its two cells `A` and `B` in loop with reversed polarities. `EMF` of each shell is `E` and internal resistance `r`. Potential difference across cell `A` or `B` is (here `n gt 4`)

Options

A. `(2E)/(n)`
B. `2E (1 - (1)/(n))`
C. `(4E)/(n)`
D. `2E (1 - (2)/(n))`

Answer

D

Explanation

(d) The two opposite cells `A` and `B` will cancel two more other cells, so net emf will be `n - 4`. So current
`I = ((n - 4)E)/(nr)`, Now p.d. across `A` or `B`
`=E + Ir` (as they will be in charging state)
`= E + ((n - 4)E)/(n) = 2 E (1 - (2)/(n))`